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Proof step 2 from proof prove-ext (previous step)

A.rows = B.rows, A.cols = B.cols, ∀ m, n. m < B.rows ∧ n < B.cols → A[m, n] = B[m, n] ⊦ A = B

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ A = B ↔ A.rows = B.rows ∧ A.cols = B.cols ∧ (∀ m, n. m < A.rows ∧ n < A.cols → A[m, n] = B[m, n])
Proof step 4:
A.rows = B.rows, A.cols = B.cols, ∀ m, n. m < B.rows ∧ n < B.cols → A[m, n] = B[m, n] ⊦ A.rows = B.rows ∧ A.cols = B.cols ∧ (∀ m, n. m < A.rows ∧ n < A.cols → A[m, n] = B[m, n])

Rule argument

spec: matrix inst: name: ext seq: ⊦ A = B ↔ A.rows = B.rows ∧ A.cols = B.cols ∧ (∀ m, n. m < A.rows ∧ n < A.cols → A[m, n] = B[m, n]) subst: [B, A] → [B, A] flags: rewrite

Additional information

Used terms: B, A